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# HW5 - WM Problem 2.75 2.75 The plastic block shown is...

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Unformatted text preview: WM Problem 2 .75 2.75 The plastic block shown is bonded to a rigid support and to a vertical plate to which a 240-kN load P is applied. Knowing that for the plastic used G = 1050 MPa, determine the deﬂection of the plate. A .— (amaze).- mx/psm‘ = 7.6x/0'3 M2 P - 240x103 N I, "'.',; ~ Y 2 0 [03 V. ’\ [if :‘ ASE’: qt!6>:(lo‘3 : ZSX/O‘ PC; 120: ' G = tOSOx/o‘ Pa A/ L 25 4 _ = g=7§5§€52 = 23.8on/o“ 50 Dimensions in mm h : SOMM "' 0.0501» S = 1'0": (o.05o\(23.2:oy/o'33 = Hawaii") Lorna“ Problem 2.80 2.80 For the elastomeric bearing in Prob. 2.79 with b = 220 mm and a = 30 mm, determine the shearing modulus G and the shear stress Tfor a maximum lateral load P = 19 kN and a maximum displacement 6 = 12 mm. Shearing ¥ome P = W x103 N Area A =‘ (200 MmXQZO m..\ ‘-' '41-!»thg My} 1‘ LWXIO'3 1m" 3 73:13: ”“93: 431.2”10‘134 A W") = 431 kPa ‘ 5 12M ,_ SLan‘xna s+raivx T: E‘ =—3—o—ﬁ — 0-400 Shean‘n 5 Mock) pus 3 G=~§-= W = Losowo‘ Pa = 1.080 MPQ. *1 Problem 2_82 2.82 Two blocks of rubber with a modulus of rigidity G = 1.50 l<si are bonded to rigid supports and to a plate AB. Knowing that b = 8 in. and c = 5 1n., determine the largest allowable load P and the smallest allowable thickness a of the blocks if the shearing stress in the rubber is not to exceed 210 psis and the deﬂection of the plate a W is to be at least% in. lp Shanna sin/was Shear M3 s‘l rm‘m Prob|em 3. 5 3.5 (0,) For the 60-mm-diameter solid cylinder and loading shown, determine the maximum shearing stress. (b) Determine the inner diameter of the hollow cylinder, of 80-mm outer diameter, for which the maximum stress is the same as in part a. (0.) Soﬁa—J sLaH: c =i—d = 'Ji(0.oco) = 0.030 m 2’ _19 _ 2T , (235000) N" J ‘ Tl'C“ TiZo.030)3 = H'7.‘893><l0" 11.; me MPQA (b) Flo/Wow sha‘H'l CL: ﬁg] ': %(0.0803= 0.010 M l 11.“ q (o 0403“- W ' 77(‘ll'7393x ID‘) | 0-032237% 39.37‘YMM Cpl : QC: = €15 mm 4 =13: x/o" m C r H \ P roblem 3' 1 3 3.13 Under normal operating conditions, the electric motor exerts a torque of 2.8 kN - m on shaﬁ AB. Knowing that each shaﬁ is solid dete ' ' ' stress in (a) shalt AB, (b) shaﬁ BC, (6) shaﬁ CD: mime the maxrmum sheanng TB = 1.4 kN - m J 48 mm T D = 0.5 kN . m _ 46 mm co.) SLAH AB: TA“ 2-2 kid-m : 2.2x/o’ N—MJ c=4zol= 23mm= 0.023.” 73MB: :9. - 2T _ (2)(2.2xto“*l J “a v (0.023)‘ = smouo‘ Pa 2;; 21-2 MP“ 4 no» SIMH BC: Tag- L4 lN-M = Mix/03 mm BM: 24“: 0.024.“ ”E 2 2T - (2M1Axlo33 _ 4 BL Trcs " ‘IT (0-01%); ‘ GL}-H7)"O PG. 13‘: €4.5Hpﬂ. “ (d SIMH co: 72,: 0.5kN-m = 0.5xzo’N-m, c-J-al- 2w.“— _. - 0.024- m 2* 2! _ (2l(o.5xloa) __ é ‘ / _ﬁ (002%)" - 73.013 KIO PC» 229:73-0MP6. 4 Problem 3 15 3.15 The solid shaﬁ shown is formed of a brass for which the allowable shearing ' stress 55 MPa. Neglecting the effect of stress concentrations, determine smallest diameters (1,4,3 and dBC for which the allowable shearing stress is not exceeded. TB=12()()N-m 7' M“ :55" M96» # 55"“)5 PA. TC = 400N111 1: ‘ TC. 2T 0" :32 17" m T. ‘ ”Tm." - m l shade AB: 7;“: moo—woo = 30° new 3] 9:" m = li-OOYlO-3M52I,OM 116553406 Minimum a“; = 2Q. ‘-'_ 42.0 mm 4 Sharpl’ BC: Tag two N-vn Cr!” (ZAZQOO) __ -3 — mx‘nt‘mtlmﬁ‘k: 26 7 33.3mm 4 Problem 3. 25 3-25 Unda normal operating eondiﬁons a motor exerts a torque ofmagnimde 2;, = 150 N » m at F. The shafts are made ofa steel for which the allowable she-ring stress is 75 ma. Knowing tint for the gears r0 = 200 mn- md r,3 — 75 mm, determine the requixed dim of (a) shaﬁ CDE, (b) m FGHV TF6 = ‘1; ~— 150 N-w‘ n; = %T=G e if; (.503 = Lfoo ”—m f“ = 75 MP4 —_— 75x/o‘ ’PQ we 3—; 3:; U 3 (a2. Sha‘Pi‘ C'DE: 3 I (13(400 ) ﬂ -3 _ c = m 7 15-03xto a... — [\$.03 ram dc”; (b) Ska-Ft FGH t C .— ﬁglggé—lfz,el) = |o_8‘lv'6’m ——' lo_8’~l-».M deg,“ = 2c. = 21-7 mm ‘- Problem 3 _ 26 3.26 Under nor-nan operating conditions a motor exerts a torque ofmagnitude T, at F. The sliaﬁs are made ofa steel felt which the allowable shearixlg stress is 85 We arxd have disuseters dab“: 22 mm and 4m x 20 mm. Knowing that rD z 150 nun and r5 = 100 mm, dekerrtxine the largest allowable value of T... . ll (X) D 'l U) ‘0 l l 1:...4 : 8§Ml>a = 85~Io‘ Pa/ Ska-Ft PG: c=gliel=lomm= o_o:g>m T}: 2—3: = ‘—Z “c317,. = g(g§ylo‘)(o-olo)"’= 133-5 N-mT‘“ SLAP-r DE: e: —,{—el=llm.... — 0.0/1". , '1'; = g ’E’Mcs = -;i_(ss'xlo‘)(o_ouﬁ3 = 177-7 N-M BY S+al+ias 7; — % 1; = +2.13)— (177-7) = ”3‘: N—m Tke lqrjesf #1:)sz Valve 0‘? T; is 4—1»; SWJCJ‘ 6F +Le #109 cdcaa'deo’ \lc/(ues. 77,: 112-5 N-m «- Problem 334 3.34 The ship atA hasjust started to drill for oil on the ocean ﬂoor at a depth of 1500 m. Knowing that the top of the ZOO-mm-diameter steel drill pipe (G = 77.2 GPa) rotates through two complete revolutions before the drill bit at B starts to operate, determine the maximum shearing stress caused in the pipe by torsion. : TL T (+qu , G: = L 1500m N = I; : GJ'QC = G_CBE_ L J“ J'L L Q : ZV‘CV T Z<Zﬁlrad 2 i2.5éé V‘aol C’Jicoz IOOMM': O-lOOM 5. -: 77.2 GPQ = 77.2%)0’ Pa 'E= (77.2wlot)(;2.5éé)(o.too) : \$4.7»I/o‘ P4 i500 ’7: = G‘L7 MPa. ‘ Problem 3.36 300 N ~ In 3.36 The electric motor exerts a 500 N - m torque on the aluminum shaﬁ ABCD when it is rotating at a constant speed. Knowing that G = 27 GPa and that the torques exerted on pulleysB and C areasshown, determine the angle twist between (a)B and :3 C. (b)BandD. 200"" V (a) An7!c'(§+wfs‘f Ee‘l’wcen 8 Mp! C t, a .. 3‘ 44 mm T“ = 200 N-m) L.,_— I. 2 m A _ 12 _ J— : : 7 ”7f?" Na 5 ' m cud 0.022.“, G 27x10 Pa. 40mm y - I L _, . 31;; 1C: - 3G7. 97 X10 W- y 1m TL. 200 [.2 t . —“ ' L M ) ‘ 24.-IS7¥IO'3NA (Pg/c: L384 q GP“: GA” ‘ (27xloquzé7ﬂ7uo‘) ‘ (M Anjlc o'F +v\$s+ between B and D. Ta, = 500 N'md Lab =03m) C =55}? 0.02.4»1, G: 27!}Dq'Pq ”\$3." = %(0-02+)7= SUJSSx/o" m" C? = (Somme) _ CID (nyloqxszmsuoﬁ) ‘ CPm = <Pe/c+ (Pa. = “Lin/63+ 3!.9soxzo“ = 56.137v/o‘3ml CPan» ‘3 3’.ch spasm/0‘3 n1 Problem 3.48 3.48 In the gear-and-shaﬂ system shown the diameter of shaﬁs are aim9 = 2 in. and dCD = 1.5 in. Knowing that G = 11.2 X 10‘ psi, determine the angle through which end D of shaﬁ CD rotates. Cajcujdhovx a? +ovztues Ci‘V‘CO W'Pcrem.{* I A! Cow +46} \$00“; lo 61‘ w c en 7 ears B (and C ‘ “a = 13\$— - :E.’ _ g 4 '7 ~11; ., . . i T = 5 kip - in. FED:- 5 “(PM T 5ND—5 Ila-in TAe = #Zﬁ‘xlog) 9 I25» ID-5 kip-in 2ft Twrs'! in shat} A81 L: 1.5% = I9 in. CT‘éﬁIAgfLOO'm. J” : EC.“ = 590.00)" 7 [.5702 m“ = fl; _ 312.5 x103 )(132 = q -3 a (PB/A GI _ CH-2XIO‘)(LS7O8) [2'78 ”0 r1 \$2.9”th A B. cp,3 = cps/A = 12.7qulo‘3‘m! TMjQAJ’ZJ Ju‘spiacemai aleecm aim/e S r VéCPs 2 WC (Dc Villa/{Ton («If C. \$37 %Cp¢ ? [—sz(12.784>‘[0-3) "' 31.973 "/0—3 F4 “my in \$1.31 CD: L s 2H: : 24 m. ~=§ ,e 0.75 In. J = EEC" .- gums)" 9 0.46170: mt CPD/c: 1 ~‘——-————~——-——(g"'°“’("‘” = 11,557,110“ wJ GJ ' (Hay IO‘XOHWol) Remit-w d D . on: cpc + cpm = 53.5sono‘3mt cpl; 3.o7° 4 ...
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HW5 - WM Problem 2.75 2.75 The plastic block shown is...

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